Point cloud registration: a mini-review of current state, challenging issues and future directions

Nathan Brightman; Lei Fan; Yang Zhao; Nathan Brightman; Lei Fan; Yang Zhao

doi:10.3934/geosci.2023005

AIMS Geosciences

2023, Volume 9, Issue 1: 68-85. doi: 10.3934/geosci.2023005

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Mini review Special Issues

Point cloud registration: a mini-review of current state, challenging issues and future directions

Department of Civil Engineering, Xi'an Jiaotong-Liverpool University, 111 Renai Road, Suzhou 215123, China

Received: 01 September 2022 Revised: 30 October 2022 Accepted: 08 December 2022 Published: 10 January 2023

A point cloud is a set of data points in space. Point cloud registration is the process of aligning two or more 3D point clouds collected from different locations of the same scene. Registration enables point cloud data to be transformed into a common coordinate system, forming an integrated dataset representing the scene surveyed. In addition to those reliant on targets being placed in the scene before data capture, there are various registration methods available that are based on using only the point cloud data captured. Until recently, cloud-to-cloud registration methods have generally been centered upon the use of a coarse-to-fine optimization strategy. The challenges and limitations inherent in this process have shaped the development of point cloud registration and the associated software tools over the past three decades. Based on the success of deep learning methods applied to imagery data, attempts at applying these approaches to point cloud datasets have received much attention. This study reviews and comments on more recent developments in point cloud registration without using any targets and explores remaining issues, based on which recommendations on potential future studies in this topic are made.

Keywords:

Citation: Nathan Brightman, Lei Fan, Yang Zhao. Point cloud registration: a mini-review of current state, challenging issues and future directions[J]. AIMS Geosciences, 2023, 9(1): 68-85. doi: 10.3934/geosci.2023005

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Abstract

1. Introduction

For a convex function $\sigma:\mathrm{I}\subseteq \bf{R}\to \bf{R}$ on $\mathrm{I}$ with $\nu_{1}, \nu_{2}\in \mathrm{I}$ and $\nu_{1} < \nu_{2}$ , the Hermite-Hadamard inequality is defined by ^[1]:

$\begin{equation} \sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right) \leq \frac{1}{\nu_{2}-\nu_{1}}\int_{\nu_{1}}^{\nu_{2}}\sigma(\eta)d\eta \leq \frac{\sigma(\nu_{1})+\sigma(\nu_{2})}{2}. \end{equation}$

(1.1)

The Hermite-Hadamard integral inequality (1.1) is one of the most famous and commonly used inequalities. The recently published papers ^[2,3,4] are focused on extending and generalizing the convexity and Hermite-Hadamard inequality.

The situation of the fractional calculus (integrals and derivatives) has won vast popularity and significance throughout the previous five decades or so, due generally to its demonstrated applications in numerous seemingly numerous and great fields of science and engineering ^[5,6,7].

Now, we recall the definitions of Riemann-Liouville fractional integrals.

Definition 1.1 (^[5,6,7]). Let $\sigma\in L_{1}[\nu_{1}, \nu_{2}].$ The Riemann-Liouville integrals $J_{\nu_{1}+}^{\vartheta }\sigma$ and $J_{\nu_{2}-}^{\vartheta }\sigma$ of order $\vartheta > 0$ with $\nu_{1}\geq 0$ are defined by

$\begin{equation} J_{\nu_{1}+}^{\vartheta }\sigma(x) = \frac{1}{\Gamma (\vartheta )}\int_{\nu_{1}}^{x}(x-\eta)^{\vartheta -1}\sigma(\eta)d\eta,\; \ \ \ \nu_{1} < x \end{equation}$

(1.2)

and

$\begin{equation} J_{\nu_{2}-}^{\vartheta }\sigma(x) = \frac{1}{\Gamma (\vartheta )}\int_{x}^{\nu_{2}}(\eta-x)^{\vartheta -1}\sigma(\eta)d\eta,\; \ \ x < \nu_{2}, \end{equation}$

(1.3)

respectively. Here $\Gamma (\vartheta)$ is the well-known Gamma function and $J_{\nu_{1}+}^{0}\sigma(x) = J_{\nu_{2}-}^{0}\sigma(x) = \sigma(x).$

With a huge application of fractional integration and Hermite-Hadamard inequality, many researchers in the field of fractional calculus extended their research to the Hermite-Hadamard inequality, including fractional integration rather than ordinary integration; for example see ^{[8,9,10,11,12,13,14,15,16,17,18,19,20,21]}.

In this paper, we consider the integral inequality of Hermite-Hadamard-Mercer type that relies on the Hermite-Hadamard and Jensen-Mercer inequalities. For this purpose, we recall the Jensen-Mercer inequality: Let $0 < x_{1}\leq x_{2}\leq \cdots\leq x_{n}$ and $\mu = (\mu _{1}, \mu_{2}, \ldots, \mu _{n})$ nonnegative weights such that $\sum_{i = 1}^{n}\mu_{i} = 1$ . Then, the Jensen inequality ^[22,23] is as follows, for a convex function $\sigma$ on the interval $[\nu_{1}, \nu_{2}]$ , we have

$\begin{equation} \sigma\Bigg(\sum\limits_{i = 1}^{n}\mu _{i}x_{i}\Bigg)\leq \sum\limits_{i = 1}^{n}\mu _{i}\sigma(x_{i}), \end{equation}$

(1.4)

where for all $x_{i}\in \lbrack \nu_{1}, \nu_{2}]$ and $\mu_{i}\in \lbrack 0, 1]$ , $(i = \overline{1, n})$ .

Theorem 1.1 (^[2,23]). If $\sigma$ is convex function on $[\nu_{1}, \nu_{2}]$ , then

$\begin{equation} \sigma\left(\nu_{1}+\nu_{2}-\sum\limits^{n}_{i = 1}\mu_{i}x_{i}\right)\leq \sigma(\nu_{1})+\sigma(\nu_{2})-\sum\limits^{n}_{i = 1}\mu_{i}\sigma(x_{i}), \end{equation}$

(1.5)

for each $x_{i}\in[\nu_{1}, \nu_{2}]$ and $\mu_{i}\in[0, 1]$ , $(i = \overline{1, n})$ with $\sum^{n}_{i = 1}\mu_{i} = 1$ . For some results related with Jensen-Mercer inequality, see ^[24,25,26].

In view of above indices, we establish new integral inequalities of Hermite-Hadamard-Mercer type for convex functions via the Riemann-Liouville fractional integrals in the current project. Particularly, we see that our results can cover the previous researches.

2. Main results

Theorem 2.1. For a convex function $\sigma:[\nu_{1}, \nu_{2}]\subseteq \bf{R} \to\bf{R}$ with $x, y\in[\nu_{1}, \nu_{2}]$ , we have:

$\begin{multline} \sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right)\leq \frac{2^{\vartheta -1} \Gamma\left(\vartheta +1\right)}{\left( y-x\right)^{\vartheta}} \Biggl[ J_{\left(\nu_{1}+\nu_{2}-y\right)+}^{\vartheta}\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \\ +\; J_{\left(\nu_{1}+\nu_{2}-x\right)-}^{\vartheta}\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right)\Biggr] \leq \sigma\left(\nu_{1}\right)+\sigma\left(\nu_{2}\right)-\frac{\sigma\left(x\right)+\sigma\left(y\right)}{2}. \end{multline}$

(2.1)

Proof. By using the convexity of $\sigma$ , we have

$\begin{equation} \sigma\left( \nu_{1}+\nu_{2}-\frac{u+v}{2}\right) \leq \frac{1}{2}\left[ \sigma\left( \nu_{1}+\nu_{2}-u\right) +\sigma\left( \nu_{1}+\nu_{2}-v\right) \right], \end{equation}$

(2.2)

and above with $u = \frac{1-\eta}{2}x+\frac{1+\eta}{2}y$ , $v = \frac{1+\eta}{2}x+\frac{1-\eta}{2}y$ , where $x, y\in[\nu_{1}, \nu_{2}]$ and $\eta\in [0, 1]$ , leads to

$\begin{multline} \sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right)\leq\frac{1}{2} \Biggl[\sigma\left(\nu_{1}+\nu_{2}-\left(\frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right)\right) \\ +\sigma\left(\nu_{1}+\nu_{2}-\left(\frac{1+\eta}{2}x+\frac{1-\eta}{2}y\right)\right)\Biggr]. \end{multline}$

(2.3)

Multiplying both sides of (2.3) by $\eta^{\vartheta -1}$ and then integrating with respect to $\eta$ over $[0, 1]$ , we get

$\begin{align*} \frac{1}{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) &\leq \frac{1}{2}\left[\int_{0}^{1}\eta^{\vartheta-1} \sigma\left(\nu_{1}+\nu_{2}-\left(\frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right)\right)d\eta\right. \\ &+\left. \int_{0}^{1}\eta^{\vartheta -1}\sigma\left( \nu_{1}+\nu_{2}-\left( \frac{1+\eta}{2}x+\frac{1-\eta}{2}y\right)\right)d\eta\right] \\ & = \frac{1}{2}\left[ \frac{2^{\vartheta }}{\left( y-x\right) ^{\vartheta }}\int_{\nu_{1}+\nu_{2}-y}^{\nu_{1}+\nu_{2}-\frac{x+y}{2}} \left( \left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right)-w\right) ^{\vartheta -1}\sigma\left( w\right) dw\right. \\ &\left. +\frac{2^{\vartheta}}{\left(y-x\right)^{\vartheta}}\int_{\nu_{1}+\nu_{2}-\frac{x+y}{2}}^{\nu_{1}+\nu_{2}-x} \left( w-\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right) ^{\vartheta-1}\sigma\left( w\right) dw\right] \\ & = \frac{2^{\vartheta-1}\Gamma \left(\vartheta\right)}{\left(y-x\right)^{\vartheta}} \left[J_{\left(\nu_{1}+\nu_{2}-y\right)+}^{\vartheta}\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right) +\; J_{\left(\nu_{1}+\nu_{2}-x\right)-}^{\vartheta}\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right], \end{align*}$

and thus the proof of first inequality in (2.1) is completed.

On the other hand, we have by using the Jensen-Mercer inequality:

$\begin{eqnarray} \sigma\left( \nu_{1}+\nu_{2}-\left( \frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right) \right) &\leq &\sigma\left( \nu_{1}\right) +\sigma\left( \nu_{2}\right) -\left( \frac{1-\eta}{2}\sigma\left( x\right) +\frac{1+\eta}{2}\sigma\left( y\right) \right) \end{eqnarray}$

(2.4)

$\begin{eqnarray} \sigma\left( \nu_{1}+\nu_{2}-\left( \frac{1+\eta}{2}x+\frac{1-\eta}{2}y\right) \right) &\leq &\sigma\left( \nu_{1}\right) +\sigma\left( \nu_{2}\right) -\left( \frac{1+\eta}{2}\sigma\left( x\right) +\frac{1-\eta}{2}\sigma\left( y\right) \right). \end{eqnarray}$

(2.5)

Adding inequalities (2.4) and (2.5) to get

$\begin{multline} \sigma\left( \nu_{1}+\nu_{2}-\left( \frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right) \right) +\sigma\left(\nu_{1}+\nu_{2}-\left( \frac{1+\eta}{2}x+\frac{1-\eta}{2}y\right) \right) \\ \leq 2\left[\sigma\left(\nu_{1}\right)+\sigma\left(\nu_{2}\right)\right]-\left[\sigma\left(x\right)+\sigma\left(y\right)\right]. \end{multline}$

(2.6)

Multiplying both sides of (2.6) by $\eta^{\vartheta -1}$ and then integrating with respect to $\eta$ over $[0, 1]$ to obtain

$\begin{multline*} \int_{0}^{1}\eta^{\vartheta -1}\sigma\left( \nu_{1}+\nu_{2}-\left( \frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right) \right)d\eta +\int_{0}^{1}\eta^{\vartheta-1}\sigma\left(\nu_{1}+\nu_{2}-\left(\frac{1+\eta}{2}x+\frac{1-\eta}{2}y\right)\right)d\eta \\ \leq \frac{2}{\vartheta}\left[\sigma\left(\nu_{1}\right)+\sigma\left(\nu_{2}\right)\right] -\frac{1}{\vartheta }\left[ \sigma\left( x\right) +\sigma\left( y\right) \right]. \end{multline*}$

By making use of change of variables and then multiplying by $\frac{\vartheta }{2}$ , we get the second inequality in (2.1).

Remark 2.1. If we choose $\vartheta = 1$ , $x = \nu_{1}$ and $y = \nu_{2}$ in Theorem 2.1, then the inequality (2.1) reduces to (1.1).

Corollary 2.1. Theorem 2.1 with

● $\vartheta = 1$ becomes [24, Theorem 2.1].

● $x = \nu_{1}$ and $y = \nu_{2}$ becomes:

$\begin{align*} \sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right)\leq\frac{2^{\vartheta-1}\Gamma\left(\vartheta+1\right)}{\left(\nu_{2}-\nu_{1}\right)^{\vartheta}} \left[J_{\nu_{1}+}^{\vartheta }\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right)+\; J_{\nu_{2}-}^{\vartheta }\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right)\right] \leq \frac{\sigma\left(\nu_{1}\right)+\sigma\left(\nu_{2}\right) }{2}, \end{align*}$

which is obtained by Mohammed and Brevik in ^[10].

The following Lemma linked with the left inequality of (2.1) is useful to obtain our next results.

Lemma 2.1. Let $\sigma:[\nu_{1}, \nu_{2}]\subseteq\bf{R} \to\bf{R}$ be a differentiable function on $(\nu_{1}, \nu_{2})$ and $\sigma'\in L[\nu_{1}, \nu_{2}]$ with $\nu_{1}\leq \nu_{2}$ and $x, y\in[\nu_{1}, \nu_{2}]$ . Then, we have:

$\begin{multline} \frac{2^{\vartheta -1}\Gamma \left( \vartheta +1\right) }{\left( y-x\right)^{\vartheta}} \left[ J_{\left( \nu_{1}+\nu_{2}-y\right) +}^{\vartheta }\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right) +\; J_{\left( \nu_{1}+\nu_{2}-x\right)-}^{\vartheta}\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right)\right] \\ -\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right) = \frac{\left(y-x\right)}{4}\int_{0}^{1}\eta^{\vartheta} \Bigl[\sigma'\left(\nu_{1}+\nu_{2}-\left(\frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right)\right) \\ -\sigma'\left(\nu_{1}+\nu_{2}-\left(\frac{1+\eta}{2}x+\frac{1-\eta}{2}y\right)\right)\Bigr]d\eta. \end{multline}$

(2.7)

Proof. From right hand side of (2.7), we set

$\begin{align} \varpi_{1}-\varpi_{2}&:= \int_{0}^{1}\eta^{\vartheta}\left[\sigma'\left(\nu_{1}+\nu_{2} -\left(\frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right)\right) -\sigma'\left( \nu_{1}+\nu_{2}-\left( \frac{1+\eta}{2}x+\frac{1-\eta}{2}y\right)\right)\right]d\eta \\ & = \int_{0}^{1}\eta^{\vartheta }\sigma'\left(\nu_{1}+\nu_{2}-\left(\frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right)\right)d\eta \\ &-\int_{0}^{1}\eta^{\vartheta}\sigma'\left(\nu_{1}+\nu_{2}-\left(\frac{1+\eta}{2}x+\frac{1-\eta}{2}y\right)\right) d\eta. \end{align}$

(2.8)

By integrating by parts with $w = \nu_{1}+\nu_{2}-\left(\frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right)$ , we can deduce:

$\begin{align*} \varpi_{1}& = -\frac{2}{\left( y-x\right) }\sigma\left(\nu_{1}+\nu_{2}-y\right)+\frac{2\vartheta }{\left(y-x\right)} \int_{0}^{1}\eta^{\vartheta -1}\sigma\left( \nu_{1}+\nu_{2}-\left( \frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right) \right) d\eta \\ & = -\frac{2}{\left(y-x\right)}\sigma\left(\nu_{1}+\nu_{2}-y\right)+\frac{2^{\vartheta+1}\vartheta}{\left(y-x\right)^{\vartheta+1}} \int_{\nu_{1}+\nu_{2}-y}^{\nu_{1}+\nu_{2}-\frac{x+y}{2}}\sigma\left(\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right)-w\right)^{\vartheta -1} \sigma\left(w\right) dw \\ & = -\frac{2}{\left(y-x\right)}\sigma\left( \nu_{1}+\nu_{2}-y\right) +\frac{2^{\vartheta+1}\Gamma\left(\vartheta+1\right)}{\left( y-x\right) ^{\vartheta +1}} J_{\left(\nu_{1}+\nu_{2}-y\right)+}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right). \end{align*}$

Similarly, we can deduce:

$\begin{equation*} \varpi_{2} = \frac{2}{y-x}\sigma\left(\nu_{1}+\nu_{2}-x\right) -\frac{2^{\vartheta+1}\Gamma\left(\vartheta+1\right)}{\left( y-x\right) ^{\vartheta +1}} J_{\left( \nu_{1}+\nu_{2}-x\right)-}^{\vartheta }\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right). \end{equation*}$

By substituting $\varpi_{1}$ and $\varpi_{2}$ in (2.8) and then multiplying by $\frac{\left(y-x\right)}{4}$ , we obtain required identity (2.7).

Corollary 2.2. Lemma 2.1 with

● $\vartheta = 1$ becomes:

$\begin{multline*} \frac{1}{y-x}\int_{\nu_{1}+\nu_{2}-y}^{\nu_{1}+\nu_{2}-x}\sigma\left(w\right) dw-\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) = \frac{\left( y-x\right) }{4}\int_{0}^{1}\eta\left[ \sigma'\left( \nu_{1}+\nu_{2}-\left( \frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right) \right) \right. \\ \left. -\sigma'\left( \nu_{1}+\nu_{2}-\left( \frac{1+\eta}{2}x+\frac{1-\eta}{2}y\right)\right) \right] d\eta. \end{multline*}$

● $\vartheta = 1$ , $x = \nu_{1}$ and $y = \nu_{2}$ becomes:

$\begin{multline*} \frac{1}{\nu_{2}-\nu_{1}}\int_{\nu_{1}}^{\nu_{2}}\sigma\left( w\right) dw-\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right) = \frac{\left( \nu_{2}-\nu_{1}\right) }{4}\int_{0}^{1}\eta \left[ \sigma'\left(\nu_{1}+\nu_{2}-\left( \frac{1-\eta}{2}\nu_{1}+\frac{1+\eta}{2}\nu_{2}\right) \right) \right. \\ \left. -\sigma'\left( \nu_{1}+\nu_{2}-\left( \frac{1+\eta}{2}\nu_{1}+\frac{1-\eta}{2}\nu_{2}\right)\right) \right] d\eta. \end{multline*}$

● $x = \nu_{1}$ and $y = \nu_{2}$ becomes:

$\begin{multline*} \frac{2^{\vartheta -1}\Gamma \left( \vartheta +1\right) }{\left( \nu_{2}-\nu_{1}\right)^{\vartheta }} \left[ J_{\nu_{1}+}^{\vartheta }\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right) +\; J_{\nu_{2}-}^{\vartheta }\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right) \right]-\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right) \\ = \frac{\left( \nu_{2}-\nu_{1}\right) }{4}\int_{0}^{1}\eta^{\vartheta } \left[\sigma'\left( \nu_{1}+\nu_{2}-\left( \frac{1-\eta}{2}\nu_{1}+\frac{1+\eta}{2}\nu_{2}\right) \right) \right. \left. -\sigma'\left( \nu_{1}+\nu_{2}-\left( \frac{1+\eta}{2}\nu_{1}+\frac{1-\eta}{2}\nu_{2}\right) \right) \right] d\eta. \end{multline*}$

Theorem 2.2. Let $\sigma:[\nu_{1}, \nu_{2}]\subseteq\bf{R}\to\bf{R}$ be a differentiable function on $(\nu_{1}, \nu_{2})$ and $|\sigma'|$ is convex on $[\nu_{1}, \nu_{2}]$ with $\nu_{1}\leq \nu_{2}$ and $x, y\in[\nu_{1}, \nu_{2}]$ . Then, we have:

$\begin{multline} \Biggl| \frac{2^{\vartheta -1}\Gamma \left( \vartheta +1\right) }{\left(y-x\right) ^{\vartheta }} \Biggl[ J_{\left( \nu_{1}+\nu_{2}-y\right) +}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) +\; J_{\left( \nu_{1}+\nu_{2}-x\right) -}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \Biggr] \\ -\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right)\Biggr| \leq \frac{\left( y-x\right) }{2\left( 1+\vartheta \right) }\left[ \left| \sigma'\left( \nu_{1}\right) \right| +\left| \sigma'\left(\nu_{2}\right) \right| -\frac{\left| \sigma'\left( x\right) \right| +\left| \sigma'\left( y\right) \right| }{2}\right]. \end{multline}$

(2.9)

Proof. By taking modulus of identity (2.7), we get

$\begin{multline*} \Biggl|\frac{2^{\vartheta-1}\Gamma\left(\vartheta+1\right)}{\left(y-x\right)^{\vartheta}} \left[J_{\left(\nu_{1}+\nu_{2}-y\right)+}^{\vartheta}\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right) +\; J_{\left( \nu_{1}+\nu_{2}-x\right) -}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right] \\ -\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \Biggr|\leq \frac{\left( y-x\right) }{4} \left[\int_{0}^{1}\eta^{\vartheta }\left|\sigma'\left( \nu_{1}+\nu_{2}-\left( \frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right) \right) \right| d\eta\right. \\ \left. +\int_{0}^{1}\eta^{\vartheta }\left| \sigma'\left( \nu_{1}+\nu_{2}-\left( \frac{1+\eta}{2}x+\frac{1-\eta}{2}y\right) \right) \right| d\eta\right]. \end{multline*}$

Then, by applying the convexity of $|\sigma'|$ and the Jensen-Mercer inequality on above inequality, we get

$\begin{align*} &\Biggl|\frac{2^{\vartheta-1}\Gamma\left(\vartheta+1\right)}{\left(y-x\right)^{\vartheta}} \left[J_{\left(\nu_{1}+\nu_{2}-y\right)+}^{\vartheta}\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right) +\; J_{\left( \nu_{1}+\nu_{2}-x\right) -}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right] \\ &-\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \Biggr|\leq \frac{\left( y-x\right) }{4}\left[ \int_{0}^{1}\eta^{\vartheta }\left[ \left| \sigma'\left( \nu_{1}\right) \right| +\left| \sigma'\left( \nu_{2}\right) \right| -\left( \frac{1+\eta}{2}\left| \sigma^{\prime }\left( x\right) \right| +\frac{1-\eta}{2}\right) \left| \sigma'\left( y\right) \right| \right] d\eta\right. \\ &\left. +\int_{0}^{1}\eta^{\vartheta }\left[ \left| \sigma'\left(\nu_{1}\right) \right| +\left| \sigma'\left( \nu_{2}\right) \right| -\left(\frac{1-\eta}{2}\left|\sigma'\left(x\right) \right| +\frac{1+\eta}{2}\right)\left| \sigma'\left( y\right)\right| \right] d\eta\right] \\ & = \frac{\left(y-x\right)}{2\left(1+\vartheta \right)}\left[ \left|\sigma'\left( \nu_{1}\right)\right| +\left|\sigma'(\nu_{2})\right|-\frac{\left|\sigma'\left(x\right)\right|+\left|\sigma'\left(y\right)\right|}{2}\right], \end{align*}$

which completes the proof of Theorem 2.2.

Corollary 2.3. Theorem 2.2 with

● $\vartheta = 1$ becomes:

$\begin{align*} \left|\frac{1}{y-x}\int_{\nu_{1}+\nu_{2}-y}^{\nu_{1}+\nu_{2}-x}\sigma(w)dw-\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right)\right| \leq \frac{\left(y-x\right)}{4}\left[\left| \sigma'\left(\nu_{1}\right)\right|+\left|\sigma'\left(\nu_{2}\right)\right| -\frac{\left|\sigma'\left(x\right)\right|+\left|\sigma'\left(y\right)\right|}{2}\right]. \end{align*}$

● $\vartheta = 1$ , $x = \nu_{1}$ and $y = \nu_{2}$ becomes [27, Theorem 2.2].

● $x = \nu_{1}$ and $y = \nu_{2}$ becomes:

$\begin{align*} \left\vert \frac{1}{\nu_{2}-\nu_{1}}\int_{\nu_{1}}^{\nu_{2}}\sigma\left( w\right) dw-\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right) \right\vert \leq \frac{\left( \nu_{2}-\nu_{1}\right) }{4}\left[ \frac{\left\vert \sigma^{\prime }\left( \nu_{1}\right) \right\vert +\left\vert \sigma^{\prime }\left( \nu_{2}\right) \right\vert }{2}\right] . \end{align*}$

Theorem 2.3. Let $\sigma:[\nu_{1}, \nu_{2}]\subseteq\bf{R}\to\bf{R}$ be a differentiable function on $(\nu_{1}, \nu_{2})$ and $\left| \sigma'\right|^{q}, q > 1$ is convex on $[\nu_{1}, \nu_{2}]$ with $\nu_{1}\leq \nu_{2}$ and $x, y\in[\nu_{1}, \nu_{2}]$ . Then, we have:

$\begin{multline} \left| \frac{2^{\vartheta -1}\Gamma \left( \vartheta +1\right) }{\left( y-x\right) ^{\vartheta }}\left[ J_{\left( \nu_{1}+\nu_{2}-y\right) +}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}- \frac{x+y}{2}\right) \right. \right. \\ \left. \left. +\; J_{\left( \nu_{1}+\nu_{2}-x\right) -}^{\vartheta }\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right] -\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right| \\ \leq \frac{\left( y-x\right) }{4\sqrt[p]{\vartheta p+1}} \left[ \left( \left| \sigma'\left( \nu_{1}\right) \right|^{q}+\left| \sigma'\left(\nu_{2}\right)\right| ^{q} -\left(\frac{\left|\sigma'(x) \right|^{q}+3\left|\sigma'(y)\right|^{q}}{4}\right)\right)^{\frac{1}{q}}\right. \\ \left. +\left( \left| \sigma'\left( \nu_{1}\right) \right|^{q}+\left|\sigma'\left(\nu_{2}\right)\right|^{q} -\left(\frac{3\left|\sigma'(x)\right|^{q}+\left|\sigma'(y)\right|^{q}}{4}\right)\right)^{\frac{1}{q}}\right], \end{multline}$

(2.10)

where $\frac{1}{p}+\frac{1}{q} = 1$ .

Proof. By taking modulus of identity (2.7) and using Hölder's inequality, we get

$\begin{multline*} \left| \frac{2^{\vartheta -1}\Gamma \left( \vartheta +1\right) }{\left(y-x\right)^{\vartheta }} \left[ J_{\left( \nu_{1}+\nu_{2}-y\right)+}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right. \right. \\ \left. \left. +\; J_{\left(\nu_{1}+\nu_{2}-x\right)-}^{\vartheta }\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right] -\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right| \\ \leq \frac{\left( y-x\right) }{4}\left( \int_{0}^{1}\eta^{\vartheta p}\right)^{\frac{1}{p}} \left\{\left(\int_{0}^{1} \left|\sigma'\left(\nu_{1}+\nu_{2}-\left(\frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right)\right)\right|^{q}d\eta\right) ^{\frac{1}{q}}\right. \\ \left. +\left(\int_{0}^{1} \left|\sigma'\left(\nu_{1}+\nu_{2}-\left(\frac{1+\eta}{2}x+\frac{1-\eta}{2}y\right)\right)\right|^{q}d\eta\right) ^{\frac{1}{q}}\right\}. \end{multline*}$

Then, by applying the Jensen-Mercer inequality with the convexity of $\left|\sigma'\right|^{q}$ , we can deduce

$\begin{align*} &\left| \frac{2^{\vartheta -1}\Gamma \left( \vartheta +1\right) }{\left(y-x\right) ^{\vartheta }} \left[J_{\left(\nu_{1}+\nu_{2}-y\right)+}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right. \right. \\ &\left. \left. +\; J_{\left( \nu_{1}+\nu_{2}-x\right) -}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right] -\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right| \\ &\leq \frac{\left( y-x\right) }{4}\left( \int_{0}^{1}\eta^{\vartheta p}\right)^{\frac{1}{p}} \left\{ \left( \int_{0}^{1}\left| \sigma'\left( \nu_{1}\right)\right| ^{q}+\left| \sigma'\left( \nu_{2}\right) \right|^{q} -\left(\frac{1-\eta}{2}\left|\sigma'(x)\right|^{q}+\frac{1+\eta}{2}\left|\sigma'(y)\right|^{q}\right)\right)^{\frac{1}{q}}\right. \\ &\left. +\left( \int_{0}^{1}\left| \sigma'\left( \nu_{1}\right)\right| ^{q}+\left| \sigma'\left( \nu_{2}\right) \right|^{q} -\left(\frac{1+\eta}{2}\left| \sigma'\left( x\right)\right|^{q}+\frac{1-\eta}{2}\left| \sigma'\left( y\right) \right|^{q} \right) \right)^{\frac{1}{q}}\right\} \\ & = \frac{\left(y-x\right)}{4\sqrt[p]{\vartheta p+1}}\left[\left( \left| \sigma'\left( \nu_{1}\right) \right| ^{q} +\left|\sigma'\left( \nu_{2}\right) \right| ^{q}-\left( \frac{\left|\sigma'\left( x\right) \right|^{q} +3\left| \sigma'\left(y\right) \right|^{q} }{4}\right) \right) ^{\frac{1}{q}}\right. \\ &\left. +\left( \left| \sigma'\left( \nu_{1}\right) \right|^{q}+\left| \sigma'\left( \nu_{2}\right) \right| ^{q} -\left(\frac{3\left|\sigma'(x)\right|^{q}+\left|\sigma'(y)\right|^{q}}{4}\right)\right)^{\frac{1}{q}}\right], \end{align*}$

which completes the proof of Theorem 2.3.

Corollary 2.4. Theorem 2.3 with

● $\vartheta = 1$ becomes:

$\begin{multline*} \left|\frac{1}{y-x}\int_{\nu_{1}+\nu_{2}-y}^{\nu_{1}+\nu_{2}-x}\sigma\left(w\right)dw -\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right)\right| \\ \leq \frac{\left( y-x\right)}{4\sqrt[p]{p+1}}\left[\left( \left| \sigma'\left( \nu_{1}\right) \right|^{q} +\left|\sigma'\left( \nu_{2}\right) \right| ^{q}-\left( \frac{\left|\sigma'\left( x\right) \right|^{q} +3\left|\sigma'\left(y\right) \right|^{q}}{4}\right)\right)^{\frac{1}{q}}\right. \\ \left. +\left(\left|\sigma'\left( \nu_{1}\right)\right|^{q}+\left| \sigma'\left( \nu_{2}\right)\right|^{q} -\left(\frac{3\left|\sigma'(x)\right|^{q}+\left|\sigma'(y)\right|^{q} }{4}\right)\right)^{\frac{1}{q}}\right]. \end{multline*}$

● $\vartheta = 1$ , $x = \nu_{1}$ and $y = \nu_{2}$ becomes:

$\begin{align*} &&\left\vert \frac{1}{\nu_{2}-\nu_{1}}\int_{\nu_{1}}^{\nu_{2}}\sigma\left( w\right) dw-\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right) \right\vert \leq \frac{\left( \nu_{2}-\nu_{1}\right) }{2^{\frac{2}{p}}}\left(\frac{1}{p+1}\right)^{\frac{1}{p}}\left[ \left\vert \sigma^{\prime }\left( \nu_{1}\right) \right\vert +\left\vert \sigma^{\prime }\left( \nu_{2}\right) \right\vert \right] . \end{align*}$

● $x = \nu_{1}$ and $y = \nu_{2}$ becomes:

$\begin{align*} &\left\vert \frac{2^{\vartheta-1}\Gamma\left(\vartheta+1\right)}{\left(\nu_{2}-\nu_{1}\right)^{\vartheta}} \left[ J_{\nu_{1}+}^{\vartheta }\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right) \right. \right. \left. \left. +\; J_{\nu_{2}-}^{\vartheta }\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right) \right] -\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right) \right\vert \\ &\leq \frac{2^{\vartheta-1-\frac{2}{q}}}{\nu_{2}-\nu_{1}}\left(\frac{1}{p+1}\right)^{\frac{1}{p}}\left[ \left\vert \sigma'\left( \nu_{1}\right) \right\vert +\left\vert \sigma'\left( \nu_{2}\right) \right\vert \right]. \end{align*}$

Theorem 2.4. Let $\sigma:[\nu_{1}, \nu_{2}]\subseteq\bf{R}\to\bf{R}$ be a differentiable function on $(\nu_{1}, \nu_{2})$ and $\left|\sigma'\right|^{q}, q\geq 1$ is convex on $[\nu_{1}, \nu_{2}]$ with $\nu_{1}\leq \nu_{2}$ and $x, y\in[\nu_{1}, \nu_{2}]$ . Then, we have:

$\begin{multline} \left| \frac{2^{\vartheta -1}\Gamma \left( \vartheta +1\right) }{\left(y-x\right) ^{\vartheta }} \left[ J_{\left( \nu_{1}+\nu_{2}-y\right) +}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right)\right.\right. \\ \left. \left. +\; J_{\left( \nu_{1}+\nu_{2}-x\right) -}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right] -\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right| \\ \leq \frac{\left(y-x\right)}{4\left(\vartheta+1\right)}\left[\left(\left|\sigma'(\nu_{1})\right|^{q} +\left|\sigma'\left(\nu_{2}\right)\right|^{q}-\left( \frac{\left| \sigma'\left(x\right)\right|^{q} +\left( 2\vartheta+3\right)\left|\sigma'(y)\right|^{q}}{2\left(\vartheta+2\right) }\right)\right)^{\frac{1}{q}}\right. \\ \left. +\left( \left|\sigma'\left(\nu_{1}\right)\right|^{q}+\left| \sigma'\left( \nu_{2}\right)\right|^{q} -\left(\frac{\left(2\vartheta+3\right)\left|\sigma'(x)\right|^{q}+\left|\sigma'(y)\right|^{q}}{2\left( \vartheta+2\right) }\right) \right) ^{\frac{1}{q}}\right]. \end{multline}$

(2.11)

Proof. By taking modulus of identity (2.7) with the well-known power mean inequality, we can deduce

$\begin{multline*} \left| \frac{2^{\vartheta -1}\Gamma \left( \vartheta +1\right) }{\left(y-x\right) ^{\vartheta }} \left[ J_{\left( \nu_{1}+\nu_{2}-y\right) +}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right. \right. \\ \left. \left. +\; J_{\left( \nu_{1}+\nu_{2}-x\right) -}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right] -\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right| \\ \leq \frac{\left( y-x\right) }{4}\left( \int_{0}^{1}\eta^{\vartheta }\right) ^{1-\frac{1}{q}} \left\{\left(\int_{0}^{1}\eta^{\vartheta} \left|\sigma'\left(\nu_{1}+\nu_{2}-\left(\frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right)\right)\right|^{q}d\eta\right)^{\frac{1}{q}}\right. \\ \left. +\left( \int_{0}^{1}\eta^{\vartheta } \left|\sigma'\left(\nu_{1}+\nu_{2}-\left(\frac{1+\eta}{2}x+\frac{1-\eta}{2}y\right)\right)\right|^{q}d\eta\right)^{\frac{1}{q}}\right\} . \end{multline*}$

By applying the Jensen-Mercer inequality with the convexity of $\left|\sigma'\right|^{q}$ , we can deduce

$\begin{align*} &\left| \frac{2^{\vartheta -1}\Gamma \left( \vartheta +1\right) }{\left(y-x\right)^{\vartheta}} \left[ J_{\left( \nu_{1}+\nu_{2}-y\right)+}^{\vartheta}\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right. \right. \\ &\left. \left. +\; J_{\left( \nu_{1}+\nu_{2}-x\right) -}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right)\right] -\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right| \\ &\leq \frac{\left( y-x\right)}{4}\left(\int_{0}^{1}\eta^{\vartheta }\right)^{1-\frac{1}{q}} \left\{ \left( \int_{0}^{1}\eta^{\vartheta }\left[ \left|\sigma'\left( \nu_{1}\right) \right|^{q} +\left| \sigma'\left(\nu_{2}\right) \right| ^{q}-\left( \frac{1-\eta}{2}\left| \sigma'\left(x\right)\right|^{q} +\frac{1+\eta}{2}\left| \sigma'\left( y\right)\right| ^{q}\right) \right] \right) ^{\frac{1}{q}}\right. \\ &\left. +\left( \int_{0}^{1}\eta^{\vartheta }\left[ \left| \sigma'\left(\nu_{1}\right)\right|^{q} +\left| \sigma'\left( \nu_{2}\right) \right|^{q}-\left( \frac{1+\eta}{2}\left| \sigma'\left( x\right) \right|^{q} +\frac{1-\eta}{2}\left| \sigma'\left( y\right) \right|^{q}\right) \right] \right) ^{\frac{1}{q}}\right\} \\ & = \frac{\left( y-x\right) }{4\left( \vartheta +1\right) }\left[\left(\left|\sigma'\left(\nu_{1}\right)\right|^{q} +\left|\sigma'\left( \nu_{2}\right) \right| ^{q}-\left( \frac{\left| \sigma'\left(x\right) \right|^{q} +\left(2\vartheta +3\right)\left|\sigma'(y)\right|^{q}}{2\left(\vartheta+2\right)}\right)\right)^{\frac{1}{q}}\right. \\ &\left. +\left( \left| \sigma'\left( \nu_{1}\right) \right|^{q}+\left|\sigma'\left( \nu_{2}\right)\right|^{q} -\left( \frac{\left( 2\vartheta +3\right) \left| \sigma'\left( x\right) \right|^{q} +\left|\sigma'\left( y\right)\right|^{q}}{2\left(\vartheta+2\right)}\right)\right)^{\frac{1}{q}}\right], \end{align*}$

which completes the proof of Theorem 2.4.

Corollary 5. Theorem 2.4 with

● $q = 1$ becomes Theorem 2.2.

● $\vartheta = 1$ becomes:

$\begin{multline*} \left| \frac{1}{y-x}\int_{\nu_{1}+\nu_{2}-y}^{\nu_{1}+\nu_{2}-x}\sigma\left(w\right) dw -\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right| \\ \leq \frac{\left( y-x\right) }{8}\left[ \left( \left|\sigma'\left( \nu_{1}\right)\right|^{q} +\left| \sigma'\left( \nu_{2}\right)\right|^{q}-\left( \frac{\left| \sigma'\left( x\right)\right|^{q} +5\left|\sigma'\left(y\right)\right|^{q}}{6}\right)\right)^{\frac{1}{q}}\right. \\ \left. +\left( \left| \sigma'\left( \nu_{1}\right) \right|^{q}+\left|\sigma'\left(\nu_{2}\right)\right|^{q} -\left(\frac{5\left|\sigma'(x)\right|^{q}+\left|\sigma'(y)\right|^{q}}{6}\right)\right)^{\frac{1}{q}}\right]. \end{multline*}$

● $\vartheta = 1$ , $x = \nu_{1}$ and $y = \nu_{2}$ becomes:

$\begin{multline*} \left| \frac{1}{\nu_{2}-\nu_{1}}\int_{\nu_{1}}^{\nu_{2}}\sigma\left( w\right) dw -\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right)\right| \\ \leq \frac{\left(y-x\right)}{8}\left[\left(\frac{5\left|\sigma'\left(\nu_{1}\right)\right|^{q} +\left| \sigma'\left( \nu_{2}\right)\right| ^{q}}{6}\right)^{\frac{1}{q}} +\left(\frac{\left|\sigma'(\nu_{1})\right|^{q}+5\left| \sigma'(\nu_{2})\right|^{q}}{6}\right)^{\frac{1}{q}}\right]. \end{multline*}$

● $x = \nu_{1}$ and $y = \nu_{2}$ becomes:

$\begin{multline*} \left| \frac{2^{\vartheta-1}\Gamma\left(\vartheta+1\right)}{\left(\nu_{2}-\nu_{1}\right)^{\vartheta}} \left[J_{\nu_{1} +}^{\vartheta}\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right)+\; J_{\nu_{2} -}^{\vartheta }\sigma\left( \frac{\nu_{1}+\nu_{2}}{2}\right)\right] -\sigma\left( \frac{\nu_{1}+\nu_{2}}{2}\right)\right| \\ \leq \frac{\left(\nu_{2}-\nu_{1}\right)}{4\left(\vartheta+1\right)} \left[\left(\frac{\left(2\vartheta+3\right)\left|\sigma'(\nu_{1})\right|^{q}+\left|\sigma'(\nu_{2})\right|^{q}}{2\left(\vartheta+2\right)}\right)^{\frac{1}{q}} +\left(\frac{\left|\sigma'(\nu_{1})\right|^{q}+\left(2\vartheta+3\right)\left|\sigma'(\nu_{2})\right|^{q}}{2\left(\vartheta+2\right)}\right) ^{\frac{1}{q}}\right]. \end{multline*}$

3. Applications to special means

Here, we consider the following special means:

● The arithmetic mean:

$\mathcal{A}(\nu_{1},\nu_{2}) = \frac{\nu_{1}+\nu_{2}}{2}, \quad \nu_{1}, \nu_{2}\geq 0.$

● The harmonic mean:

$\mathcal{H}(\nu_{1},\nu_{2}) = \frac{2\nu_{1}\nu_{2}}{\nu_{1}+\nu_{2}}, \quad \nu_{1}, \nu_{2} > 0.$

● The logarithmic mean:

$\begin{equation*} \mathcal{L}(\nu_{1},\nu_{2}) = \begin{cases} \frac{\nu_{2}-\nu_{1}}{\ln \nu_{2}-\ln \nu_{1}}, & \text{if}\; \nu_{1}\neq \nu_{2}, \\ \nu_{1}, & \text{if}\; \nu_{1} = \nu_{2}, \end{cases} \quad \nu_{1}, \nu_{2} > 0. \end{equation*}$

● The generalized logarithmic mean:

$\begin{equation*} \mathcal{L}_{n}(\nu_{1},\nu_{2}) = \begin{cases} \left[\frac{\nu_{2}^{n+1}-\nu_{1}^{n+1}}{\left(n+1\right)(\nu_{2}-\nu_{1})}\right]^{\frac{1}{n}}, & \text{if}\; \nu_{1}\neq \nu_{2} \\ \nu_{1}, & \text{if}\; \nu_{1} = \nu_{2}, \end{cases} \quad \nu_{1},\nu_{2} > 0;\; n\in\bf{Z}\setminus \{-1,0\}. \end{equation*}$

Proposition 3.1. Let $0 < \nu_{1} < \nu_{2}$ and $n\in\bf{N}$ , $n \geq 2$ . Then, for all $x, y\in[\nu_{1}, \nu_{2}]$ , we have:

(3.1)

Proof. By applying Corollary 2.3 (first item) for the convex function $\sigma(x) = x^{n}, \, x > 0$ , one can obtain the result directly.

Proposition 3.2. Let $0 < \nu_{1} < \nu_{2}$ . Then, for all $x, y\in[\nu_{1}, \nu_{2}]$ , we have:

(3.2)

Proof. By applying Corollary 2.3 (first item) for the convex function $\sigma(x) = \frac{1}{x}, \, x > 0$ , one can obtain the result directly.

Proposition 3.3. Let $0 < \nu_{1} < \nu_{2}$ and $n\in\bf{N}$ , $n \geq 2$ . Then, we have:

$\begin{equation} \left| \mathcal{L}_{n}^{n}(\nu_{1},\nu_{2}) -\mathcal{A}^{n}(\nu_{1},\nu_{2}) \right| \leq \frac{n(\nu_{2}-\nu_{1})}{4} \left[ \mathcal{A}\left(\nu_{1}^{n-1}, \nu_{2}^{n-1}\right) \right], \end{equation}$

(3.3)

and

$\begin{equation} \left| \mathcal{L}^{-1}(\nu_{1},\nu_{2}) -\mathcal{A}^{-1}(\nu_{1},\nu_{2}) \right| \leq \frac{(\nu_{2}-\nu_{1})}{4}\mathcal{H}^{-1}\left( \nu_{1}^{2},\nu_{2}^{2}\right) . \end{equation}$

(3.4)

Proof. By setting $x = \nu_{1}$ and $y = \nu_{2}$ in results of Proposition 3.1 and Proposition 3.2, one can obtain the Proposition 3.3.

Proposition 3.4. Let $0 < \nu_{1} < \nu_{2}$ and $n\in\bf{N}$ , $n \geq 2$ . Then, for $q > 1, \, \frac{1}{p}+\frac{1}{q} = 1$ and for all $x, y\in[\nu_{1}, \nu_{2}],$ we have:

$\begin{multline} \left| \mathcal{L}_{n}^{n}\left( \nu_{1}+\nu_{2}-y,\nu_{1}+\nu_{2}-x\right) -\left( 2\mathcal{A}(\nu_{1},\nu_{2}) -\mathcal{A}\left( x,y\right) \right) ^{n}\right| \\ \leq \frac{n(y-x)}{4\sqrt[p]{p+1}}\Bigg\{\left[ 2\mathcal{A}\left( \nu_{1}^{q(n-1)}, \nu_{2}^{q(n-1)}\right) -\frac{1}{2}\mathcal{A}\left(x^{q(n-1)}, 3y^{q(n-1)}\right) \right]^{\frac{1}{q}} \\ +\left[ 2\mathcal{A}\left( \nu_{1}^{q(n-1)}, \nu_{2}^{q(n-1)}\right) -\frac{1}{2}\mathcal{A}\left(3x^{q(n-1)}, y^{q(n-1)}\right) \right]^{\frac{1}{q}}\Bigg\}. \end{multline}$

(3.5)

Proof. By applying Corollary 2.4 (first item) for convex function $\sigma\left(x\right) = x^{n}, \, x > 0$ , one can obtain the result directly.

Proposition 3.5. Let $0 < \nu_{1} < \nu_{2}$ . Then, for $q > 1, \, \frac{1}{p}+\frac{1}{q} = 1$ and for all $x, y\in[\nu_{1}, \nu_{2}]$ , we have:

$\begin{multline} \left| \mathcal{L}^{-1}\left( \nu_{1}+\nu_{2}-y,\nu_{1}+\nu_{2}-x\right) -\left( 2\mathcal{A}(\nu_{1},\nu_{2}) -\mathcal{A}\left(x,y\right)\right)^{-1}\right| \\ \leq \frac{\sqrt[q]{2}(y-x)}{4\sqrt[p]{p+1}}\Bigg\{\left[ \mathcal{H}^{-1}\left(\nu_{1}^{2q},\nu_{2}^{2q}\right) -\frac{3}{4}\mathcal{H}^{-1}\left(x^{2q}, 3y^{2q}\right) \right]^{\frac{1}{q}} \\ +\left[\mathcal{H}^{-1}\left(\nu_{1}^{2q},\nu_{2}^{2q}\right) -\frac{3}{4}\mathcal{H}^{-1}\left(3x^{2q}, y^{2q}\right) \right]^{\frac{1}{q}}\Bigg\}. \end{multline}$

(3.6)

Proof. By applying Corollary 2.4 (first item) for the convex function $\sigma\left(x\right) = \frac{1}{x}, \, x > 0$ , one can obtain the result directly.

Proposition 3.6. Let $0 < \nu_{1} < \nu_{2}$ and $n\in\bf{N}$ , $n \geq 2$ . Then, for $q > 1$ and $\frac{1}{p}+\frac{1}{q} = 1$ , we have:

$\begin{multline} \left| \mathcal{L}_{n}^{n}(\nu_{1},\nu_{2}) -\mathcal{A}^{n}(\nu_{1},\nu_{2}) \right| \leq \frac{n(\nu_{2}-\nu_{1})}{4\sqrt[p]{p+1}} \Bigg\{\left[ 2\mathcal{A}\left( \nu_{1}^{q(n-1)}, \nu_{2}^{q(n-1)}\right) -\frac{1}{2}\mathcal{A}\left( \nu_{1}^{q(n-1)}, 3\nu_{2}^{q(n-1)}\right) \right]^{\frac{1}{q}} \\ +\left[ 2\mathcal{A}\left( \nu_{1}^{q(n-1)}, \nu_{2}^{q(n-1)}\right) -\frac{1}{2}\mathcal{A}\left(3\nu_{1}^{q(n-1)}, \nu_{2}^{q(n-1)}\right) \right]^{\frac{1}{q}}\Bigg\}, \end{multline}$

(3.7)

and

$\begin{multline} \left| \mathcal{L}^{-1}(\nu_{1},\nu_{2}) -\mathcal{A}^{-1}(\nu_{1},\nu_{2}) \right| \leq \frac{\sqrt[q]{2}(\nu_{2}-\nu_{1})}{4\sqrt[p]{p+1}} \Bigg\{\left[ \mathcal{H}^{-1}\left( \nu_{1}^{2q}, \nu_{2}^{2q}\right) -\frac{3}{4}\mathcal{H}^{-1}\left(\nu_{1}^{2q}, 3\nu_{2}^{2q}\right) \right]^{\frac{1}{q}} \\ +\left[ \mathcal{H}^{-1}\left( \nu_{1}^{2q}, \nu_{2}^{2q}\right) -\frac{3}{4}\mathcal{H}^{-1}\left(3\nu_{1}^{2q}, \nu_{2}^{2q}\right) \right]^{\frac{1}{q}}\Bigg\}. \end{multline}$

(3.8)

Proof. By setting $x = \nu_{1}$ and $y = \nu_{2}$ in results of Proposition 3.4 and Proposition 3.5, one can obtain the Proposition 3.6.

4. Conclusions

As we emphasized in the introduction, integral inequality is the most important field of mathematical analysis and fractional calculus. By using the well-known Jensen-Mercer and power mean inequalities, we have proved new inequalities of Hermite-Hadamard-Mercer type involving Riemann-Liouville fractional operators. In the last section, we have considered some propositions in the context of special functions; these confirm the efficiency of our results.

Acknowledgements

We would like to express our special thanks to the editor and referees. Also, the first author would like to thank Prince Sultan University for funding this work through research group Nonlinear Analysis Methods in Applied Mathematics (NAMAM) group number RG-DES-2017-01-17.

Conflict of interest

The authors declare no conflict of interest.

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